Elementary symmetric polynomial task with three variables Can anyone help me to wite this as sum or product of elementary symmetric polynomial.
$$\frac xy+\frac yx +\frac xz + \frac zx +\frac yz + \frac zy =7$$

I tried to set under one fraction, but I didn't go much far.
 A: \begin{align*}
  yz(y+z)+zx(z+x)+xy(x+y) &= 7xyz  \\
  yz(x+y+z)+zx(x+y+z)+xy(x+y+z)-3xyz &= 7xyz \\
  (x+y+z)(yz+zx+xy) &= 10xyz
\end{align*}
A: First, we get rid of the fractions. We get $$x^2 y+x^2 z+x y^2+x z^2+y^2 z+y z^2= 7xyz$$
Now we need to express the left hand side in terms of elementary symmetric polynomials. Since it is third-degree, we try computing $(x+y+z)^3$ and we get $$x^3+y^3+z^3 +3x^2 y+3 x^2 z+3 x y^2+6 x y z+3 x z^2+3 y^2 z+3 y z^2$$ We see that, if we get rid of the terms $x^3+y^3+z^3$ and $6xyz$, we have what we want. So, computing $$\tfrac13\left((x+y+z)^3-6xyz-(x^3+y^3+z^3)\right)=x^2 y+x^2 z+x y^2+x z^2+y^2 z+y z^2$$ which is what we wanted. So the expression becomes $$\tfrac13\left((x+y+z)^3-6xyz-(x^3+y^3+z^3)\right)=7xyz$$
or, simplified:
$$(x+y+z)^3-(x^3+y^3+z^3)=27xyz$$
Hope this helped!
A: Hint: Expand  $(x+y+z)(xy+yz+zx)$.  You will be very close to the "numerator" you mentioned in a comment. (Since comments can disappear,   we add that this numerator is obtained by multiplying the expression of the OP by $xyz$.)
A: In the LHS of
$$x^2 y+x^2 z+x y^2+x z^2+y^2 z+y z^2= 7xyz$$
you recognize all the binary terms from the development of $(x+y+z)^3$. To complete the cube,
$$(x+y+z)^3=\color{blue}{x^3+y^3+z^3 + 3}(x^2 y+x^2 z+x y^2+x z^2+y^2 z+y z^2)\color{blue}{+6xyz}\\= x^3+y^3+z^3 +27xyz.$$
